Acceleration Vector in Circular Motion

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SUMMARY

The discussion focuses on calculating the acceleration vector for a go-cart moving on a circular track with a radius of 40 meters. The go-cart accelerates from rest to 60 km/hr in 20 seconds, leading to a tangential acceleration of 0.8333 m/s². The correct formula for the magnitude of the acceleration vector is established as mag of a = √(a_t² + (R * α)²), where α is the angular acceleration. The final calculated magnitude of the acceleration vector is confirmed to be 33.34 m/s², although the initial calculations contained errors regarding the relationship between tangential and angular acceleration.

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  • Understanding of circular motion dynamics
  • Familiarity with tangential and angular acceleration concepts
  • Knowledge of basic kinematics equations
  • Ability to convert units (e.g., km/hr to m/s)
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  • Study the relationship between tangential and centripetal acceleration in circular motion
  • Learn how to derive angular acceleration from tangential acceleration
  • Explore the implications of acceleration vectors in non-linear motion
  • Practice problems involving circular motion with varying speeds
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mindarson
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Homework Statement



Consider a go-cart moving on a circular track of radius R = 40 m. Suppose it starts from rest and speeds up to 60 km/hr in 20 seconds, with a constant rate of increase of the speed.
Calculate the magnitude of a.



Homework Equations



a(t) = a_t + α

where the first term is the tangential acceleration and the second is the angular acceleration.


The Attempt at a Solution



I begin with the magnitude of a:

mag of a = [(R^2)(α^2) - (R^2)(ω^4)]^.5

Then I calculate ω:

ω = v/R = (60000m/3600s)/40m = .4167 /s

And I can calculate the tangential acceleration:

a_t = Δv/Δt = (60000m/3600s)/20s = .8333 m/s^2

Now I can calculate dω/dt = α:

α = r*a_t = 40 m * .8333 m/s^2 = 33.33 /s^2

Now to calculate the magnitude of the acceleration vector:

mag of a = [(a_t)^2 + α^2]^.5 = 33.34 m/s^2

But apparently this is wrong. Can anyone point out what I've missed? Seems pretty straightforward but I just can't get it.
 
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mindarson said:

Homework Statement



Consider a go-cart moving on a circular track of radius R = 40 m. Suppose it starts from rest and speeds up to 60 km/hr in 20 seconds, with a constant rate of increase of the speed.
Calculate the magnitude of a.

Homework Equations



a(t) = a_t + α

where the first term is the tangential acceleration and the second is the angular acceleration.
[/b]

Did you mean centripetal acceleration instead of angular one?

mindarson said:
Now I can calculate dω/dt = α:
α = r*a_t = 40 m * .8333 m/s^2 = 33.33 /s^2

This is wrong. the tangential acceleration is R times the angular acceleration a_t=αR.ehild
 
Last edited:

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